ГДЗ по алгебре 11 класс Никольский Параграф 8. Уравнения-следствия Задание 39

\[\boxed{\mathbf{39.}}\]

\[\textbf{а)}\log_{3}\sqrt{x} + \log_{3}{\sqrt{x + 8} =}1\]

\[\log_{3}{\sqrt{x}\sqrt{x + 8}} = \log_{3}3\]

\[x > 0;\]

\[\sqrt{x}\sqrt{x + 8} = 3\]

\[x(x + 8) = 9\]

\[x^{2} + 8x - 9 = 0\]

\[D_{1} = 16 + 9 = 25\]

\[x_{1} = - 4 + 5 = 1;\]

\[x_{2} = - 4 - 5 =\]

\[= - 9 < 0\ (не\ корень).\]

\[Ответ:x = 1.\]

\[\textbf{б)}\log_{2}\sqrt{x + 3} + \log_{2}\sqrt{x + 6} = 1\]

\[\log_{2}{\sqrt{x + 3}\sqrt{x + 6}} = \log_{2}2\]

\[x > - 3.\]

\[\sqrt{x + 3}\sqrt{x + 6} = 2\]

\[(x + 3)(x + 6) = 4\]

\[x^{2} + 3x + 6x + 18 - 4 = 0\]

\[x^{2} + 9x + 14 = 0\]

\[x_{1} + x_{2} = - 9;\ \ x_{1} \cdot x_{2} = 14\]

\[x_{1} = - 7 < - 3\ (не\ корень);\ \ \]

\[x + 2 = - 2.\]

\[Ответ:x = - 2.\]

\[\textbf{в)}\log_{3}\sqrt{x + 4} + \log_{3}\sqrt{x - 4} = 1\]

\[\log_{3}{\sqrt{x + 4}\sqrt{x - 4}} = \log_{3}3\]

\[x > 4.\]

\[\sqrt{x + 4}\sqrt{x - 4} = 3\]

\[(x + 4)(x - 4) = 9\]

\[x^{2} - 16 = 9\]

\[x^{2} = 25\]

\[x_{1} = - 5 < 4\ (не\ корень);\]

\[x_{2} = 5.\]

\[Ответ:x = 5.\]

\[\textbf{г)}\log_{4}\sqrt{x + 3} + \log_{4}\sqrt{x - 3} = 1\]

\[\log_{4}{\sqrt{x + 3}\sqrt{x - 3}} = \log_{4}4\]

\[x > 3.\]

\[\sqrt{x + 3}\sqrt{x - 3} = 4\]

\[(x + 3)(x - 3) = 16\]

\[x^{2} - 9 = 16\]

\[x^{2} = 25\]

\[x_{1} = - 5 < 3\ (не\ корень);\]

\[x_{2} = 5.\]

\[Ответ:x = 5.\]